FZU 1901 KMP找前后缀等串

Period II
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
FZU
1901

Description

For each prefix with length P of a given string S,if
S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

Input

Input contains multiple cases.
The first line contains an integer T representing the number of cases. Then following T cases.

Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.

Output

For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending order.

Sample Input

4oooacmacmacmacmacmafzufzufzufstostootssto

Sample Output

Case #1: 31 2 3Case #2: 63 6 9 12 15 16Case #3: 43 6 9 10Case #4: 29 12

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
using namespace std;
int nex[1000005];
int k;
int ans[1000005];
void pre(string str){
nex[0]=-1;
int i=0,j=-1;
while(i<str.size()){
if(j==-1||str[i]==str[j]){
j++;i++;
nex[i]=j;
}else
j=nex[j];
}
}
void kmp(string str){
pre(str);
int len=str.size();
int i=len-1;
k=0;
while(i>-1){
if(str[i]==str[len-1]) ans[k++]=len-i-1;
i=nex[i];
}
}
int main()
{
std::cout.sync_with_stdio(false);
string str;
int n;
while(cin>>n){
for(int t=1;t<=n;t++){
cin>>str;
kmp(str);
cout<<"Case #"<<t<<": "<<k<<endl;
for(int i=1;i<k;i++)
cout<<ans[i]<<' ';
cout<<str.size()<<endl;
}
}
}